Algebraic proof that the monoid ring of a torsion-free monoid is reduced

Question

In what follows, I say that a monoid $M$ is torsion-free if the $n$-th power map is injective for all $n \geq 1$. I have a proof of the following result:

Claim: if $M$ is a torsion-free commutative monoid and $k$ is a field, then $k[M]$ is reduced.

Without going into details, here's how my proof works:

• If $k$ is of characteristic $p$ and $x = \sum_{m \in M} \lambda_m m$ is a nilpotent element, then $x^{p^r} = 0$ for some $r \geq 0$, i.e.: $$ \sum_{m\in M} \lambda_m^{p^r} m^{p^r} = 0 $$ Since the $p^r$-th power map is injective, the elements $m^{p^r}$ are distinct and thus linearly independent. Hence $\lambda_m^{p^r} = 0$ for all $m \in M$, thus $\lambda_m = 0$ for all $m \in M$ and finally $x=0$.

• In characteristic zero, this follows from the following fact: let $F$ be a finite subset of $M$, then the property "every element $x \in \mathrm{Span}(F)$ which satisfies $x^2 = 0$ in $K[M]$ is zero" is expressible as a first-order property of the base field $K$. By Łos's theorem and the case of characteristic $p$, any ultraproduct of $\overline{\mathbb{F}_p}$ is an algebraically closed field of characteristic zero which satisfies these properties for all finite $F \subseteq M$. Since $ACF_0$ is complete, this is true for $\bar k$ too and thus $\bar k [M]$ is reduced. If $x^2=0$ in $k[M]$, then $x^2=0$ in $\bar k[M]$ too and thus $x=0$. So $k[M]$ is reduced.

I am bothered by the use of model theory, which seems unnatural since this is a purely algebraic statement. The reason why I took that route is the following: to prove that a sum of elements is not nilpotent is tricky because there may be unexpected cancellations caused by the cross terms. In characteristic $p$ (when $M$ is commutative), this is not a problem since all these extra terms disappear. Nevertheless, I feel like the result is simple enough that people with absolutely no knowledge of model theory would have been able to prove it.

Questions: How would other algebraists have proved the claim? Is the distinction between nonzero and zero characteristic necessary? Can we use algebraic methods instead of Łos's theorem? Can we get rid of the "commutative" hypothesis?

Thanks for any help!

Answer 1

Reduction mod $p$. Let ${\rm char}\,k=0$. If the assertion is false, then it is false over a finitely generated subring $k_0\subseteq k$. Since $k_0$ is a Jacobson ring it contains a maximal ideal $\mathfrak m$ such that $x$ is non-zero mod $\mathfrak m$. Since $k_0/\mathfrak m$ is a finite field you get a contradiction to the first part.

The argument is well-known, btw. Long time ago, I found it in a text book from the 50s or 60s, I think.